SOLUTION: Hello, I do not understand completely how to find the direction of vectors. I understand that you are suppose to use this formula: θ=arctan(y/x). What I don't understand is

Algebra ->  Vectors -> SOLUTION: Hello, I do not understand completely how to find the direction of vectors. I understand that you are suppose to use this formula: θ=arctan(y/x). What I don't understand is      Log On


   

Question 755269: Hello,
I do not understand completely how to find the direction of vectors. I understand that you are suppose to use this formula: θ=arctan(y/x). What I don't understand is what to do if you have negative y's and/or x's, and how to solve for the different quadrants. My problem that I am really confused on is this one:
What is the magnitude and direction angle of the vector <2,-sqrt(3)>?
Here is how I tried to solve it:
Magnitude = sqrt%28x%5E2%2By%5E2%29
Magnitude = sqrt%282%5E2%2B-sqrt%283%29%5E2%29
Magnitude = sqrt%284%2B3%29 = sqrt%287%29
Direction angle = arctan(y/x)
=arctan%28-sqrt%283%29%2F2%29
How would I finish solving this? Do I include the negative sign for all vectors?
Thanks!

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Think of your vector as the arrow from point O to point A:

Point A is in the fourth quadrant.
It is at a distance sqrt%287%29 from O.
The coordinates of A are
x=2=sqrt%287%29%2Acos%28theta%29
y=-sqrt%283%29=sqrt%287%29%2Asin%28theta%29
Your direction angle is
theta=arctan%28-sqrt%283%29%2F2%29=-40.9%5Eo
The angle theta=arctan%28y%2Fx%29 does not help you figure out the quadrant because <-2,sqrt(3)> would give you the same angle.
The signs of {{x}}} and y tell you the quadrant.
<(2,-sqrt(3)> and <(-2,sqrt(3)> are mirror images of each other.
<(-2,sqrt(3)> forms a 180%5Eo-40.9%5Eo=139.1%5Eo with the positive x-axis.